Derivatives of Trigonometric Functions

[Pages:7]Derivatives of Trigonometric Functions

The basic trigonometric limit:

Theorem:

lim sin x x0 x

=1=

lim

x0

x sin

x

(x in radians)

Note: In calculus, unless otherwise noted, all angles are measured in radians, and not in degrees. This theorem is sometimes referred to as the small-angle approximation because it really says that, for very small angles x, sin x x.

Note:

Cosine behaves even better near 0, where

lim cos

x0

x =1.

ex. Show that

lim cos x -1 = 0

x 0

x

lim

x 0

cos x -1 x

=

lim cos x x0 x

-1 cos x +1 cos x +1

=

lim

x0

cos2 x -1 x(cos x +1)

=

lim

x0

- sin 2 x x(cos x +1)

=

lim

x 0

-

sin x

x

lim x0

sin x cos x + 1

=

-

lim

x0

sin x

x

lim x0

sin x cos x + 1

=

-

(1) 1

0 +

1

=

0

ex.

Evaluate

lim

x0

sin 2x 5x

lim

x0

sin 2x 5x

=

1 5

lim

x0

sin 2x x

2 2

=

2 5

lim

x0

sin 2x 2x

The idea above is to match the angle in the sine function with the denominator. We'll then apply the basic trigonometric limit. To do so, first we substitute = 2x. Note that as x approaches 0, so does . Hence,

2 5

lim

x 0

sin 2x 2x

=

2 5

lim

0

sin

=

2 5

1

=

2 5

ex.

Evaluate

lim

x0

sin sin

4x 3x

lim

x0

sin 4x sin 3x

=

lim

x0

sin 4x sin 3x

x x

=

lim

x0

sin 4x x

x sin 3x

=

lim

x0

sin 4x x

lim x0

x sin 3x

Repeat the same trick as in the previous example, let = 4x and = 3x. Both and approach 0 when x does. Then apply the theorem twice.

=

lim

x0

sin 4x x

4 4

lim x0

x sin 3x

3 3

=

4 3

lim

x0

sin 4x 4x

lim x0

3x sin 3x

=

4 3

lim

0

sin

lim 0

sin

=

4 3

1

1

=

4 3

In fact, after doing a few examples like those, we can see a (very nice) pattern. To sum it up:

Suppose m and n are nonzero real numbers, then

lim sin mx = m

x0 nx

n

lim

x0

mx sin nx

=

m n

lim

x0

sin mx sin nx

=

m n

(Trivially, we also have:

lim

x0

mx nx

=

m n

.)

ex.

Evaluate

lim

x0

tan 7x 2x

lim

x0

tan 7x 2x

=

lim

x0

1 2x

sin 7x cos 7x

=

1 2

lim

x0

sin 7x x

1 cos

x

=

1 2

lim

x0

sin 7x x

lim x0

1 cos

x

= 1 7 1 = 7 211 2

Recall that since cos x is continuous everywhere, the direct substitution property applies, therefore,

lim

x 0

1 cos

x

=

1 lim cos

x

=

1 cos 0

=

1 1

=1

x0

Now, the main topic --

Derivatives of Trigonometric Functions

ex. What is the derivative of sin x?

Start with the limit definition of derivative:

d sin x = lim sin(x + h) - sin x = lim [sin x cos h + sin h cos x] - sin x

dx

h0

h

h0

h

= lim sin x cos h - sin x + lim sin h cos x = lim sin x(cos h -1) + lim sin h cos x

h0

h

h0

h

h0

h

h0 h

= lim sin x lim cos h -1 + lim sin h lim cos x = sin x (0) + (1) cos x = cos x

h0

h0

h

h h0

h0

Therefore,

d dx

sin

x

=

cos x

ex. Find the derivative of csc x.

d dx

csc

x

=

d dx

1 sin

x

=

sin

x

d 1 -1 d dx dx

(sin x)2

sin

x

=

sin

x

0 -1 cos sin 2 x

x

=

- cos x sin 2 x

=

-

1 sin

x

cos x sin x

=

-

csc

x cot

x

Therefore,

d dx

csc x

=

- csc

x cot

x

The complete list of derivatives of trigonometric functions:

1.

d sin x = cos x dx

2.

d cos x = - sin x dx

3.

d tan x = sec2 x dx

4.

d sec x = sec x tan x dx

5.

d cot x = - csc2 x dx

6.

d csc x = - csc x cot x dx

ex. Differentiate f (x) = sec x + 5 csc x f (x) = sec x tan x + 5( -csc x cot x) = sec x tan x - 5 csc x cot x

ex. Differentiate f (x) = x2 cos x - 2x sin x - 3 cos x f (x) = [x2(-sin x) + (2x) cos x] - 2[x (cos x) + (1)sin x] - 3(-sin x) = - x2 sin x + 2x cos x - 2x cos x - 2sin x + 3sin x = - x2 sin x + sin x

ex. Differentiate

s (t )

=

1

sin t - cos

t

s(t )

=

(1 -

cos

t)(cos t) - (sin t)(0 (1 - cos t)2

-

(- sin

t ))

= cos t - cos2 t - sin 2 t = cos t - (cos2 t + sin 2 t)

(1 - cos t)2

(1 - cos t)2

= cos t - 1 = - (1 - cos t) = -1 = 1 (1 - cos t)2 (1 - cos t)2 1 - cos t cos t - 1

ex. Simple Harmonic Motion Suppose the oscillating motion (in meters) of a weight attached to a spring is described by the displacement function

s(t) = 2 cos t + sin t Find its velocity and acceleration functions, and its speed and acceleration at t = /2 sec.

Velocity: v(t) = s(t) = -2 sin t + cos t Acceleration: a(t) = v(t) = -2 cos t - sin t Its speed when t = /2 is v(/2) = -2 sin (/2) + cos (/2) = -2 + 0 = 2 (m/sec) Its acceleration at the same time is a(/2) = -2 cos (/2) - sin (/2) = 0 - 1 = -1 (m/sec2)

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